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anonymous

  • one year ago

What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6? f(x) =negative one over six times x squared plus four over three times x plus eleven over six f(x) =one over six times x squared minus four over three times x minus eleven over six f(x) = three fiftieth x squared plus eleven twenty fifth x plus three fifth f(x) = two fiftieth x squared plus eleven twenty fifth x plus two fifth

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  1. anonymous
    • one year ago
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    @welshfella @nincompoop @Nnesha @ganeshie8

  2. anonymous
    • one year ago
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    @ASAAD123

  3. Nnesha
    • one year ago
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    first we should draw a point at (4,-3) and horizontal line y=-6 |dw:1442578792483:dw| this one is better

  4. anonymous
    • one year ago
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    |dw:1442578889393:dw|

  5. Nnesha
    • one year ago
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    |dw:1442578826480:dw| yes that's right :=)

  6. Nnesha
    • one year ago
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    brb wait a mint plz

  7. Nnesha
    • one year ago
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    sorry about alright so now we should pick a point to find distance between the point and focus and distance between the point and directrix you can pick any point |dw:1442579344198:dw|

  8. Nnesha
    • one year ago
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    equation would be like this \[\huge\rm \sqrt{(x-a)^2+(y-b)^2} =\sqrt{ (y-c)^2}\] c = directrix (a,b) is a focus point (4,-3) now substitute a ,b c for their values

  9. anonymous
    • one year ago
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    square root 4-a^2+-3-b=square root-3-c^2

  10. Nnesha
    • one year ago
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    well (x,y) is a point that we picked we supposed taht (a,b) is the focus point so replace a with 4 b with -3 and c with -6 c is directix

  11. anonymous
    • one year ago
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    4-4^2+-3-(-3)=square root-3-(-6)^2 like this?

  12. Nnesha
    • one year ago
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    keep the x and y variable \[\large\rm \sqrt{(x-4)^2+(y-(-3))^2} =\sqrt{ (y-(-6))^2}\] like this now take square both sides to cancel out the square root

  13. anonymous
    • one year ago
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    X^-16+Y+9=Y+36

  14. anonymous
    • one year ago
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    X^2

  15. anonymous
    • one year ago
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    is that right

  16. Nnesha
    • one year ago
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    hmm no (x-4)^2 = ?

  17. anonymous
    • one year ago
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    X^2+16

  18. Nnesha
    • one year ago
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    hmm no (x-4)^2 is same as (x-4)(x-4) foil it signs are the same in the parentheses so you should get 3 terms

  19. anonymous
    • one year ago
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    X^2-8x+16

  20. Nnesha
    • one year ago
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    yes right \[x^2-8x+16+(y-(-3))^2=(y-(-6))^2\] foil (y-(-3))^2 and then right side

  21. Nnesha
    • one year ago
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    here is a hint you will get two terms when you have (a-b)(a+b) opposite signs

  22. anonymous
    • one year ago
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    Y^2-9

  23. Nnesha
    • one year ago
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    hmm no (y-(-3))^2 first distribute -1(-3) ?

  24. anonymous
    • one year ago
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    thats 3

  25. Nnesha
    • one year ago
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    yes so (y+3)^2 is same as (y+3)(y+3)

  26. anonymous
    • one year ago
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    oh

  27. anonymous
    • one year ago
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    y^2+6y-9

  28. anonymous
    • one year ago
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    +9

  29. Nnesha
    • one year ago
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    yes right \[\large\rm x^2-8x+16+y^2+6y+9=(y-(-6))^2\] foil right side

  30. anonymous
    • one year ago
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    Y^2+12y+36

  31. anonymous
    • one year ago
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    i dont mean to rush but i have to go soon

  32. Nnesha
    • one year ago
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    yes right \[\large\rm x^2-8x+16+y^2+6y+9=y^2+12y+36\] foil right side now simplify combine like terms

  33. Nnesha
    • one year ago
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    well it will takes some time we need to apply the complete the square method .....

  34. Nnesha
    • one year ago
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    \[\large\rm x^2-8x+\color{red}{16}+\color{blue}{y^2}+6y+\color{red}{9}=\color{blue}{y^2}+12y+36\] combine like terms at left side and then move y^2 to the right side let me know what you get

  35. anonymous
    • one year ago
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    srry but i have to go thank you any ways

  36. Nnesha
    • one year ago
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    okay we will finish this whnevr u hve time

  37. princesssleelee
    • one year ago
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    @Nnesha i have this question as well that i need assistance with.

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